# How much work does it take to raise a 6 kg  weight 21 m ?

Dec 20, 2016

$W \approx 1235 J$

#### Explanation:

Work is given by $W = F \Delta r \cos \left(\theta\right)$, where $F$ is the applied force, $\Delta r$ is the displacement, and $\theta$ is the angle between the force and displacement vectors. Because the weight is moving upward and this is the direction that the force of the lift is applied, $\theta = {0}^{o}$, $W$ is positive, and $W = F \Delta y$

The force that must be exerted to lift the weight is its physical weight, given by ${F}_{g} = m g$. We are given $6 k g$ as a mass quantity, and $g$ is the free-fall acceleration, $9.8 \frac{m}{s} ^ 2$.

$F = \left(6 k g\right) \left(9.8 \frac{m}{s} ^ 2\right) = 58.8 N$

Therefore, a force of $58.8 N$ was exerted on the weight over a vertical distance of $21 m$, and

$W = 58.8 N \cdot 21 m$

$W = 1234.8 J \approx 1235 J$

In scientific notation for significant figures, $W \approx 1 \cdot {10}^{3} J$.

Note that this is the work done by the lift, not the net work done on the weight.